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If I1 is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and I2 is the moment of inertia of ring about an axis perpendicular to the plane of ring and passing through its centre formed by bending the rod, then $\frac{I_{1}}{I_{2}} =$

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$\frac{3}{π^{2}}$

$\frac{2}{π^{2}}$

$\frac{π^{2}}{3}$

$\frac{π^{2}}{2}$

answer is C.

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Detailed Solution

length of the rod $l = 2 π r r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{12}$ and $I_{2} = m r^{2} = m {(\frac{l}{2 π})}^{2} = \frac{m l^{2}}{4 π^{2}}$

$\frac{I_{1}}{I_{2}} = \frac{4 π^{2}}{12} = \frac{π^{2}}{3}$

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